Solve Radical Equations
-$$\sqrt { x + 6 } + \sqrt { 2 - x } = 4$$
A) $\{ - 2 \}$
B) $\{ 2 , - 2 \}$
C) $\{ 0 \}$
D) $\{ \sqrt { 31 } , - 2 \}$

Question

Quizplus · Accepted Answer

To solve the equation $\sqrt{x + 6} + \sqrt{2 - x} = 4$, we first isolate one of the square roots and then square both sides to eliminate the square root, following these steps:1. $\sqrt{x + 6} = 4 - \sqrt{2 - x}$2. Squaring both sides: $(\sqrt{x + 6})^2 = (4 - \sqrt{2 - x})^2$3. Simplifying: $x + 6 = 16 - 8\sqrt{2 - x} + (2 - x)$4. Rearranging: $8\sqrt{2 - x} = 12 - x$5. Squaring both sides again to eliminate the square root: $64(2 - x) = (12 - x)^2$6. Expanding and simplifying: $128 - 64x = 144 - 24x + x^2$7. Rearranging: $x^2 + 40x - 16 = 0$8. Factoring or using the quadratic formula to find $x$.However, the correct approach after isolating and squaring both sides involves careful algebraic manipulation and checking for extraneous solutions, which can arise when squaring both sides of an equation. The correct solution to the equation is $x = -2$, which satisfies the original equation when substituted back in, making sure both square roots are defined and the equation balances. Other proposed solutions do not satisfy the original equation when checked.

Solve Radical Equations - x+6+2−x=4\sqrt { x + 6 } + \sqrt { 2 - x } = 4x+6​+2−x​=4

Solve Radical Equations
- $\sqrt { x + 6 } + \sqrt { 2 - x } = 4$